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If I want to evaluate the flux of a vector field over an unfinished square (basically over three sides of a square), instead of parametrezing the 3 sides and computing the flux 3 times, can I use Green's theorem over the full square and subtract from it the flux over the line integral of the side that was not available before?

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After re-reading I think you seem to suggest \begin{align} \int\limits_\sqsubset A \cdot dr &= \int\limits_{\square = \partial S} A \cdot dr - \int\limits_{.\vert} A \cdot dr \\ &= \int\limits_S\text{curl } A \cdot dS - \int\limits_{.\vert} A \cdot dr \\ &= \int\limits_S (\partial_1 A_2 - \partial_2 A_1) \lVert dS \rVert- \int\limits_{.\vert} A \cdot dr \\ \end{align} where we extended to three dimensions in between by giving the square surface a unit normal in positive 3-direction: $S = \lVert S \rVert e_3$ and using Kelvin-Stokes instead of Green.

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  • $\begingroup$ I don't really understand your answer. Why do I need to find a vector potential A? $\endgroup$ – Omrane Dec 11 '15 at 13:19
  • $\begingroup$ Let me rephrase maybe I didn't explain myself very well: Imagine A(1,-1), B(1,1), C(-1,1), D(-1,-1) 4 vertices of the square ABCD. Now I want the flux over the curve that goes from A to B, B to C and C to D. Can I use Green's Theorem over ABCD and subtract from that the flux over CD? $\endgroup$ – Omrane Dec 11 '15 at 13:23
  • $\begingroup$ For the flux-divergence or normal form Green's Theorem, it relates the partial derivative of M with respect to x (i component of vector field) with partial derivative of N with respect to y (j component of vector field) which we add and take the integral over the region formed by the curves. $\endgroup$ – Omrane Dec 11 '15 at 13:31
  • $\begingroup$ Ok it seems we have a mismatch of dimensions. I gave the 3D version. $\endgroup$ – mvw Dec 11 '15 at 13:32
  • $\begingroup$ So is this technique of application correct? Edit: Oh I see. For now we are only using it in 2 dimensions. $\endgroup$ – Omrane Dec 11 '15 at 13:33

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